right triangles and trigonometry homework 2

Right Triangle Trigonometry Calculator

Basics of trigonometry, right triangles trigonometry calculations, example of right triangle trigonometry calculations with steps, more trigonometry and right triangles calculators (and not only).

The right triangle trigonometry calculator can help you with problems where angles and triangles meet: keep reading to find out:

• The basics of trigonometry;
• How to calculate a right triangle with trigonometry;
• A worked example of how to use trigonometry to calculate a right triangle with steps;

And much more!

Trigonometry is a branch of mathematics that relates angles to the length of specific segments . We identify multiple trigonometric functions: sine, cosine, and tangent, for example. They all take an angle as their argument, returning the measure of a length associated with the angle itself. Using a trigonometric circle , we can identify some of the trigonometric functions and their relationship with angles.

As you can see from the picture, sine and cosine equal the projection of the radius on the axis, while the tangent lies outside the circle. If you look closely, you can identify a right triangle using the elements we introduced above: let's discover the relationship between trigonometric functions and this shape.

Consider an acute angle in the trigonometric circle above: notice how you can build a right triangle where:

• The radius is the hypotenuse; and
• The sine and cosine are the catheti of the triangle.

α \alpha α is one of the acute angles, while the right angle lies at the intersection of the catheti (sine and cosine)

Let this sink in for a moment: the length of the cathetus opposite from the angle α \alpha α is its sine , sin ⁡ ( α ) \sin(\alpha) sin ( α ) ! You just found an easy and quick way to calculate the angles and sides of a right triangle using trigonometry.

The complete relationships between angles and sides of a right triangle need to contain a scaling factor, usually the radius (the hypotenuse). Identify the opposite and adjacent . We can then write:

By switching the roles of the legs, you can find the values of the trigonometric functions for the other angle.

Taking the inverse of the trigonometric functions , you can find the values of the acute angles in any right triangle.

Using the three equations above and a combination of sides, angles, or other quantities, you can solve any right triangle . The cases we implemented in our calculator are:

• Solving the triangle knowing two sides ;
• Solving the triangle knowing one angle and one side ; and
• Solving the triangle knowing the area and one side .

Take a right triangle with hypotenuse c = 5 c = 5 c = 5 and an angle α = 38 ° \alpha=38\degree α = 38° . Surprisingly enough, this is enough data to fully solve the right triangle! Follow these steps:

• Calculate the third angle: β = 90 ° − α \beta = 90\degree - \alpha β = 90° − α .
• sin ⁡ ( α ) = 0.61567 \sin(\alpha) = 0.61567 sin ( α ) = 0.61567 .
• o p p o s i t e = sin ⁡ ( α ) ⋅ h y p o t e n u s e = 0.61567 ⋅ 5 = 3.078 \mathrm{opposite} = \sin(\alpha)\cdot\mathrm{hypotenuse} = 0.61567 \cdot 5 = 3.078 opposite = sin ( α ) ⋅ hypotenuse = 0.61567 ⋅ 5 = 3.078 .
• a d j a c e n t = 0.788 ⋅ 5 = 3.94 \mathrm{adjacent} = 0.788\cdot 5 = 3.94 adjacent = 0.788 ⋅ 5 = 3.94 .

If you liked our right triangle trigonometry calculator, why not try our other related tools? Here they are:

• The trigonometry calculator ;
• The cosine triangle calculator ;
• The sine triangle calculator ;
• The trig triangle calculator ;
• The trig calculator ;
• The sine cosine tangent calculator ;
• The tangent ratio calculator ; and
• The tangent angle calculator .

How do I apply trigonometry to a right triangle?

To apply trigonometry to a right triangle, remember that sine and cosine correspond to the legs of a right triangle . To solve a right triangle using trigonometry:

• sin(α) = opposite/hypotenuse ; and
• cos(α) = adjacent/hypotenuse .
• By taking the inverse trigonometric functions , we can find the value of the angle α .
• You can repeat the procedure for the other angle.

What is the hypotenuse of a triangle with α = 30° and opposite leg a = 3?

The length of the hypotenuse is 6 . To find this result:

• Calculate the sine of α : sin(α) = sin(30°) = 1/2 .
• Apply the following formula: sin(α) = opposite/hypotenuse hypotenuse = opposite/sin(α) = 3 · 2 = 6 .

Can I apply right-triangle trigonometric rules in a non-right triangle?

Not directly: to apply the relationships between trigonometric functions and sides of a triangle, divide the shape alongside one of the heights lying inside it. This way, you can split the triangle into two right triangles and, with the right combination of data, solve it!

Adjoint matrix

Helium balloons, schwarzschild radius, sine cosine tangent.

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7.2 Right Triangle Trigonometry

Learning objectives.

In this section you will:

• Use right triangles to evaluate trigonometric functions.
• Find function values for 30° ( π 6 ) , 45° ( π 4 ) , 30° ( π 6 ) , 45° ( π 4 ) , and 60° ( π 3 ) . 60° ( π 3 ) .
• Use equal cofunctions of complementary angles.
• Use the definitions of trigonometric functions of any angle.
• Use right-triangle trigonometry to solve applied problems.

Mt. Everest, which straddles the border between China and Nepal, is the tallest mountain in the world. Measuring its height is no easy task. In fact, the actual measurement has been a source of controversy for hundreds of years. The measurement process involves the use of triangles and a branch of mathematics known as trigonometry. In this section, we will define a new group of functions known as trigonometric functions, and find out how they can be used to measure heights, such as those of the tallest mountains.

Using Right Triangles to Evaluate Trigonometric Functions

Figure 1 shows a right triangle with a vertical side of length y y and a horizontal side has length x . x . Notice that the triangle is inscribed in a circle of radius 1. Such a circle, with a center at the origin and a radius of 1, is known as a unit circle .

We can define the trigonometric functions in terms an angle t and the lengths of the sides of the triangle. The adjacent side is the side closest to the angle, x . (Adjacent means “next to.”) The opposite side is the side across from the angle, y . The hypotenuse is the side of the triangle opposite the right angle, 1. These sides are labeled in Figure 2 .

Given a right triangle with an acute angle of t , t , the first three trigonometric functions are listed.

A common mnemonic for remembering these relationships is SohCahToa, formed from the first letters of “ underline S end underline ine is underline o end underline pposite over underline h end underline ypotenuse, underline C end underline osine is underline a end underline djacent over underline h end underline ypotenuse, underline T end underline angent is underline o end underline pposite over underline a end underline djacent.”

For the triangle shown in Figure 1 , we have the following.

Given the side lengths of a right triangle and one of the acute angles, find the sine, cosine, and tangent of that angle.

• Find the sine as the ratio of the opposite side to the hypotenuse.
• Find the cosine as the ratio of the adjacent side to the hypotenuse.
• Find the tangent as the ratio of the opposite side to the adjacent side.

Evaluating a Trigonometric Function of a Right Triangle

Given the triangle shown in Figure 3 , find the value of cos α . cos α .

The side adjacent to the angle is 15, and the hypotenuse of the triangle is 17.

Given the triangle shown in Figure 4 , find the value of sin t . sin t .

Reciprocal Functions

In addition to sine, cosine, and tangent, there are three more functions. These too are defined in terms of the sides of the triangle.

Take another look at these definitions. These functions are the reciprocals of the first three functions.

When working with right triangles, keep in mind that the same rules apply regardless of the orientation of the triangle. In fact, we can evaluate the six trigonometric functions of either of the two acute angles in the triangle in Figure 5 . The side opposite one acute angle is the side adjacent to the other acute angle, and vice versa.

Many problems ask for all six trigonometric functions for a given angle in a triangle. A possible strategy to use is to find the sine, cosine, and tangent of the angles first. Then, find the other trigonometric functions easily using the reciprocals.

Given the side lengths of a right triangle, evaluate the six trigonometric functions of one of the acute angles.

• If needed, draw the right triangle and label the angle provided.
• Identify the angle, the adjacent side, the side opposite the angle, and the hypotenuse of the right triangle.
• sine as the ratio of the opposite side to the hypotenuse
• cosine as the ratio of the adjacent side to the hypotenuse
• tangent as the ratio of the opposite side to the adjacent side
• secant as the ratio of the hypotenuse to the adjacent side
• cosecant as the ratio of the hypotenuse to the opposite side
• cotangent as the ratio of the adjacent side to the opposite side

Evaluating Trigonometric Functions of Angles Not in Standard Position

Using the triangle shown in Figure 6 , evaluate sin α , cos α , tan α , sec α , csc α , and cot α . sin α , cos α , tan α , sec α , csc α , and cot α .

Another approach would have been to find sine, cosine, and tangent first. Then find their reciprocals to determine the other functions.

Using the triangle shown in Figure 7 ,evaluate sin t , cos t , tan t , sec t , csc t , and cot t . sin t , cos t , tan t , sec t , csc t , and cot t .

Finding Trigonometric Functions of Special Angles Using Side Lengths

It is helpful to evaluate the trigonometric functions as they relate to the special angles—multiples of 30° , 60° , 30° , 60° , and 45° . 45° . Remember, however, that when dealing with right triangles, we are limited to angles between 0°  and 90° . 0°  and 90° .

Suppose we have a 30° , 60° , 90° 30° , 60° , 90° triangle, which can also be described as a π 6 , π 3 , π 2 π 6 , π 3 , π 2 triangle. The sides have lengths in the relation s , 3 s , 2 s . s , 3 s , 2 s . The sides of a 45° , 45° , 90° 45° , 45° , 90° triangle, which can also be described as a π 4 , π 4 , π 2 π 4 , π 4 , π 2 triangle, have lengths in the relation s , s , 2 s . s , s , 2 s . These relations are shown in Figure 8 .

We can then use the ratios of the side lengths to evaluate trigonometric functions of special angles.

Given trigonometric functions of a special angle, evaluate using side lengths.

• Use the side lengths shown in Figure 8 for the special angle you wish to evaluate.
• Use the ratio of side lengths appropriate to the function you wish to evaluate.

Evaluating Trigonometric Functions of Special Angles Using Side Lengths

Find the exact value of the trigonometric functions of π 3 , π 3 , using side lengths.

Find the exact value of the trigonometric functions of π 4 , π 4 , using side lengths.

Using Equal Cofunction of Complements

If we look more closely at the relationship between the sine and cosine of the special angles, we notice a pattern. In a right triangle with angles of π 6 π 6 and π 3 , π 3 , we see that the sine of π 3 , π 3 , namely 3 2 , 3 2 , is also the cosine of π 6 , π 6 , while the sine of π 6 , π 6 , namely 1 2 , 1 2 , is also the cosine of π 3 . π 3 .

See Figure 9 .

This result should not be surprising because, as we see from Figure 9 , the side opposite the angle of π 3 π 3 is also the side adjacent to π 6 , π 6 , so sin ( π 3 ) sin ( π 3 ) and cos ( π 6 ) cos ( π 6 ) are exactly the same ratio of the same two sides, 3 s 3 s and 2 s . 2 s . Similarly, cos ( π 3 ) cos ( π 3 ) and sin ( π 6 ) sin ( π 6 ) are also the same ratio using the same two sides, s s and 2 s . 2 s .

The interrelationship between the sines and cosines of π 6 π 6 and π 3 π 3 also holds for the two acute angles in any right triangle, since in every case, the ratio of the same two sides would constitute the sine of one angle and the cosine of the other. Since the three angles of a triangle add to π , π , and the right angle is π 2 , π 2 , the remaining two angles must also add up to π 2 . π 2 . That means that a right triangle can be formed with any two angles that add to π 2 π 2 —in other words, any two complementary angles. So we may state a cofunction identity : If any two angles are complementary, the sine of one is the cosine of the other, and vice versa. This identity is illustrated in Figure 10 .

Using this identity, we can state without calculating, for instance, that the sine of π 12 π 12 equals the cosine of 5 π 12 , 5 π 12 , and that the sine of 5 π 12 5 π 12 equals the cosine of π 12 . π 12 . We can also state that if, for a given angle t , cos t = 5 13 , t , cos t = 5 13 , then sin ( π 2 − t ) = 5 13 sin ( π 2 − t ) = 5 13 as well.

Cofunction Identities

The cofunction identities in radians are listed in Table 1 .

Given the sine and cosine of an angle, find the sine or cosine of its complement.

• To find the sine of the complementary angle, find the cosine of the original angle.
• To find the cosine of the complementary angle, find the sine of the original angle.

Using Cofunction Identities

If sin t = 5 12 , sin t = 5 12 , find cos ( π 2 − t ) . cos ( π 2 − t ) .

According to the cofunction identities for sine and cosine, we have the following.

If csc ( π 6 ) = 2 , csc ( π 6 ) = 2 , find sec ( π 3 ) . sec ( π 3 ) .

Using Trigonometric Functions

In previous examples, we evaluated the sine and cosine in triangles where we knew all three sides. But the real power of right-triangle trigonometry emerges when we look at triangles in which we know an angle but do not know all the sides.

Given a right triangle, the length of one side, and the measure of one acute angle, find the remaining sides.

• For each side, select the trigonometric function that has the unknown side as either the numerator or the denominator. The known side will in turn be the denominator or the numerator.
• Write an equation setting the function value of the known angle equal to the ratio of the corresponding sides.
• Using the value of the trigonometric function and the known side length, solve for the missing side length.

Finding Missing Side Lengths Using Trigonometric Ratios

Find the unknown sides of the triangle in Figure 11 .

We know the angle and the opposite side, so we can use the tangent to find the adjacent side.

We rearrange to solve for a . a .

We can use the sine to find the hypotenuse.

Again, we rearrange to solve for c . c .

A right triangle has one angle of π 3 π 3 and a hypotenuse of 20. Find the unknown sides and angle of the triangle.

Using Right Triangle Trigonometry to Solve Applied Problems

Right-triangle trigonometry has many practical applications. For example, the ability to compute the lengths of sides of a triangle makes it possible to find the height of a tall object without climbing to the top or having to extend a tape measure along its height. We do so by measuring a distance from the base of the object to a point on the ground some distance away, where we can look up to the top of the tall object at an angle. The angle of elevation of an object above an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. The right triangle this position creates has sides that represent the unknown height, the measured distance from the base, and the angled line of sight from the ground to the top of the object. Knowing the measured distance to the base of the object and the angle of the line of sight, we can use trigonometric functions to calculate the unknown height.

Similarly, we can form a triangle from the top of a tall object by looking downward. The angle of depression of an object below an observer relative to the observer is the angle between the horizontal and the line from the object to the observer's eye. See Figure 12 .

Given a tall object, measure its height indirectly.

• Make a sketch of the problem situation to keep track of known and unknown information.
• Lay out a measured distance from the base of the object to a point where the top of the object is clearly visible.
• At the other end of the measured distance, look up to the top of the object. Measure the angle the line of sight makes with the horizontal.
• Write an equation relating the unknown height, the measured distance, and the tangent of the angle of the line of sight.
• Solve the equation for the unknown height.

Measuring a Distance Indirectly

To find the height of a tree, a person walks to a point 30 feet from the base of the tree. She measures an angle of 57° 57° between a line of sight to the top of the tree and the ground, as shown in Figure 13 . Find the height of the tree.

We know that the angle of elevation is 57° 57° and the adjacent side is 30 ft long. The opposite side is the unknown height.

The trigonometric function relating the side opposite to an angle and the side adjacent to the angle is the tangent. So we will state our information in terms of the tangent of 57° , 57° , letting h h be the unknown height.

The tree is approximately 46 feet tall.

How long a ladder is needed to reach a windowsill 50 feet above the ground if the ladder rests against the building making an angle of 5 π 12 5 π 12 with the ground? Round to the nearest foot.

Access these online resources for additional instruction and practice with right triangle trigonometry.

• Finding Trig Functions on Calculator
• Finding Trig Functions Using a Right Triangle
• Relate Trig Functions to Sides of a Right Triangle
• Determine Six Trig Functions from a Triangle
• Determine Length of Right Triangle Side

7.2 Section Exercises

For the given right triangle, label the adjacent side, opposite side, and hypotenuse for the indicated angle.

When a right triangle with a hypotenuse of 1 is placed in a circle of radius 1, which sides of the triangle correspond to the x - and y -coordinates?

The tangent of an angle compares which sides of the right triangle?

What is the relationship between the two acute angles in a right triangle?

Explain the cofunction identity.

For the following exercises, use cofunctions of complementary angles.

cos ( 34° ) = sin ( ___° ) cos ( 34° ) = sin ( ___° )

cos ( π 3 ) = sin ( ___ ) cos ( π 3 ) = sin ( ___ )

csc ( 21° ) = sec ( ___° ) csc ( 21° ) = sec ( ___° )

tan ( π 4 ) = cot ( ___ ) tan ( π 4 ) = cot ( ___ )

For the following exercises, find the lengths of the missing sides if side a a is opposite angle A , A , side b b is opposite angle B , B , and side c c is the hypotenuse.

cos B = 4 5 , a = 10 cos B = 4 5 , a = 10

sin B = 1 2 , a = 20 sin B = 1 2 , a = 20

tan A = 5 12 , b = 6 tan A = 5 12 , b = 6

tan A = 100 , b = 100 tan A = 100 , b = 100

sin B = 1 3 , a = 2 sin B = 1 3 , a = 2

a = 5 , ∡ A = 60° a = 5 , ∡ A = 60°

c = 12 , ∡ A = 45° c = 12 , ∡ A = 45°

For the following exercises, use Figure 14 to evaluate each trigonometric function of angle A . A .

sin A sin A

cos A cos A

tan A tan A

csc A csc A

sec A sec A

cot A cot A

For the following exercises, use Figure 15 to evaluate each trigonometric function of angle A . A .

For the following exercises, solve for the unknown sides of the given triangle.

For the following exercises, use a calculator to find the length of each side to four decimal places.

b = 15 , ∡ B = 15° b = 15 , ∡ B = 15°

c = 200 , ∡ B = 5° c = 200 , ∡ B = 5°

c = 50 , ∡ B = 21° c = 50 , ∡ B = 21°

a = 30 , ∡ A = 27° a = 30 , ∡ A = 27°

b = 3.5 , ∡ A = 78° b = 3.5 , ∡ A = 78°

Find x . x .

A radio tower is located 400 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 36° , 36° , and that the angle of depression to the bottom of the tower is 23° . 23° . How tall is the tower?

A radio tower is located 325 feet from a building. From a window in the building, a person determines that the angle of elevation to the top of the tower is 43° , 43° , and that the angle of depression to the bottom of the tower is 31° . 31° . How tall is the tower?

A 200-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 15° , 15° , and that the angle of depression to the bottom of the monument is 2° . 2° . How far is the person from the monument?

A 400-foot tall monument is located in the distance. From a window in a building, a person determines that the angle of elevation to the top of the monument is 18° , 18° , and that the angle of depression to the bottom of the monument is 3° . 3° . How far is the person from the monument?

There is an antenna on the top of a building. From a location 300 feet from the base of the building, the angle of elevation to the top of the building is measured to be 40° . 40° . From the same location, the angle of elevation to the top of the antenna is measured to be 43° . 43° . Find the height of the antenna.

There is lightning rod on the top of a building. From a location 500 feet from the base of the building, the angle of elevation to the top of the building is measured to be 36° . 36° . From the same location, the angle of elevation to the top of the lightning rod is measured to be 38° . 38° . Find the height of the lightning rod.

Real-World Applications

A 33-ft ladder leans against a building so that the angle between the ground and the ladder is 80° . 80° . How high does the ladder reach up the side of the building?

A 23-ft ladder leans against a building so that the angle between the ground and the ladder is 80° . 80° . How high does the ladder reach up the side of the building?

The angle of elevation to the top of a building in Charlotte is found to be 9 degrees from the ground at a distance of 1 mile from the base of the building. Using this information, find the height of the building.

The angle of elevation to the top of a building in Seattle is found to be 2 degrees from the ground at a distance of 2 miles from the base of the building. Using this information, find the height of the building.

Assuming that a 370-foot tall giant redwood grows vertically, if I walk a certain distance from the tree and measure the angle of elevation to the top of the tree to be 60° , 60° , how far from the base of the tree am I?

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Chapter 2: Trigonometric Ratios

Exercises: 2.2 Right Triangle Trigonometry

Exercises homework 2.2.

• Use measurements to calculate the trigonometric ratios for acute angles #1-10, 57-60
• Use trigonometric ratios to find unknown sides of right triangles #11-26
• Solve problems using trigonometric ratios #27-34, 41-46
• Use trig ratios to write equations relating the sides of a right triangle #35-40
• Use relationships among the trigonometric ratios #47-56, 61-68

Suggested Homework Problems

Here are two right triangles with a [latex]65 °[/latex] angle.

• Measure the sides [latex]AB[/latex] and [latex]BC[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
• Measure the sides [latex]AD[/latex] and [latex]DE[/latex] with a ruler. Use the lengths to estimate [latex]\sin 65°{.}[/latex]
• Use your calculator to look up [latex]\sin 65°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 1 to calculate two estimates each for the cosine and tangent of [latex]65 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\cos 65°[/latex] and [latex]\tan 65°{.}[/latex]

Here are two right triangles with a [latex]40 °[/latex] angle.

• Measure the sides [latex]AB[/latex] and [latex]AC[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
• Measure the sides [latex]AD[/latex] and [latex]AE[/latex] with a ruler. Use the lengths to estimate [latex]\cos 40°{.}[/latex]
• Use your calculator to look up [latex]\cos 40°{.}[/latex] Compare your answers. How close were your estimates?

Use the figure in Problem 2 to calculate two estimates each for the cosine and tangent of [latex]40 °{.}[/latex] Compare your estimates to your calculator’s values for [latex]\sin 40°[/latex] and [latex]\tan 40°{.}[/latex]

Exercise Group

For the right triangles in Problems 5–10,

• Find the length of the unknown side.
• Find the sine, cosine, and tangent of [latex]\theta \text{.}[/latex] Round your answers to four decimal places.

For Problems 11–16,

• Sketch and label the sides of a right triangle with angle [latex]\theta\text{.}[/latex]
• Sketch and label another right triangle with angle [latex]\theta[/latex] and longer sides.

[latex]\cos \theta = \dfrac{3}{5}[/latex]

[latex]\tan \theta = \dfrac{7}{2}[/latex]

[latex]\tan \theta = \dfrac{11}{4}[/latex]

[latex]\sin \theta = \dfrac{4}{9}[/latex]

[latex]\sin \theta = \dfrac{1}{9}[/latex]

[latex]\cos \theta = \dfrac{7}{8}[/latex]

For Problems 17–22, use one of the three trigonometric ratios to find the unknown side of the triangle. Round your answer to hundredths.

For Problems 23–26, sketch and label a right triangle with the given properties.

One angle is [latex]40°{,}[/latex] the side opposite that angle is 8 inches

One angle is [latex]65°{,}[/latex] the side adjacent to that angle is 30 yards

One angle is [latex]28°{,}[/latex] the hypotenuse is 56 feet

One leg is [latex]15[/latex] meters, the hypotenuse is [latex]18[/latex] meters

For Problems 27–34,

• Sketch a right triangle that illustrates the situation. Label your sketch with the given information.
• Choose the appropriate trig ratio and write an equation, then solve the problem.

To measure the height of cloud cover, airport controllers fix a searchlight to shine a vertical beam on the clouds. The searchlight is [latex]120[/latex] yards from the office. A technician in the office measures the angle of elevation to the light on the cloud cover at [latex]54.8°{.}[/latex] What is the height of the cloud cover?

To measure the distance across a canyon, Evel first sights an interesting rock directly opposite on the other side. He then walks [latex]200[/latex] yards down the rim of the canyon and sights the rock again, this time at an angle of [latex]18.5°[/latex] from the canyon rim. What is the width of the canyon?

A salvage ship is searching for the wreck of a pirate vessel on the ocean floor. Using sonar, they locate the wreck at an angle of depression of [latex]36.2°{.}[/latex] The depth of the ocean at their location is [latex]260[/latex] feet. How far should they move so that they are directly above the wrecked vessel?

Ramps for wheelchairs should be no steeper than an angle of [latex]6°{.}[/latex] How much horizontal distance should be allowed for a ramp that rises [latex]5[/latex] feet in height?

The radio signal from a weather balloon indicates that it is [latex]1500[/latex] meters from a meteorologist on the ground. The angle of elevation to the balloon is [latex]48°{.}[/latex] What is the balloon’s altitude?

According to Chinese legend, around 200 BC, the general Han Xin used a kite to determine the distance from his location to an enemy palace. He then dug a secret tunnel which emerged inside the palace. When the kite was directly above the palace, its angle of elevation was [latex]27°[/latex] and the string to the kite was [latex]1850[/latex] feet long. How far did Han Xin’s troops have to dig?

A cable car on a ski lift traverses a horizontal distance of [latex]1800[/latex] meters at an angle of [latex]38°{.}[/latex] How long is the cable?

Zelda is building the loft on her summer cottage. At its central point, the height of the loft is [latex]8[/latex] feet, and the pitch of the roof should be [latex]24°{.}[/latex] How long should the rafters be?

For Problems 35–40, use a trig ratio to write an equation for [latex]x[/latex] in terms of [latex]\theta{.}[/latex]

For Problems 41–44, find the altitude of the triangle. Round your answer to two decimal places.

For Problems 45 and 46, find the length of the chord [latex]AB{.}[/latex] Round your answer to two decimal places.

For Problems 47–50, fill in the table.

• In each of the figures for Problems 47-50, what is the relationship between the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex]
• Study the tables for Problems 47-50. What do you notice about the values of sine and cosine for the angles [latex]\theta[/latex] and [latex]\phi{?}[/latex] Explain why this is true.

There is a relationship between the tangent, the sine, and the cosine of any angle. Study the tables for Problems 47-50 to discover this relationship. Write your answer as an equation.

• Use the figure to explain what happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases, and why.
• Use the figure to explain what happens to [latex]\cos \theta[/latex] as [latex]\theta[/latex] increases, and why.
• What happens to [latex]\tan \theta[/latex] as [latex]\theta[/latex] increases?
• What value does your calculator give for [latex]\tan 90°{?}[/latex] Why?

Explain why it makes sense that [latex]\sin 0° = 0[/latex] and [latex]\sin 90° = 1{.}[/latex] Use a figure to illustrate your explanation.

Explain why it makes sense that [latex]\cos 0° = 1[/latex] and [latex]\cos 90° = 0{.}[/latex] Use a figure to illustrate your explanation

For Problems 57–60, explain why the trigonometric ratio is not correct.

[latex]\sin \theta = \dfrac{5}{9}[/latex]

[latex]\tan \theta = \dfrac{4}{7}[/latex]

[latex]\cos \theta = \dfrac{21}{20}[/latex]

[latex]\sin \theta = \dfrac{8}{10}[/latex]

For Problems 61–64, sketch and label a right triangle, then fill in the blank.

If [latex]\sin \theta = 0.2358\text{,}[/latex]  then [latex]\cos (90^{o} - \theta)=\underline{\qquad} \text{,}[/latex]

• If [latex]\cos \alpha = \dfrac{3}{11} \text{,}[/latex] then [latex]\underline{\qquad} (90° - \alpha) = \dfrac{3}{11}\text{.}[/latex]
• If [latex]\sin 42° = n\text{,}[/latex] then [latex]\cos \underline{\qquad} = n\text{.}[/latex]
• If [latex]\cos 13° = z\text{,}[/latex] then [latex]\sin \underline{\qquad} = z\text{.}[/latex]
• If [latex]\cos \beta = \dfrac{2}{\sqrt{7}}{,}[/latex] then [latex]\sin (90° - \beta) =\underline{\qquad} {.}[/latex]
• If [latex]\sin \phi = 0.693{,}[/latex] then [latex](90° - \phi) = 0.693{.}[/latex]
• If [latex]\cos 87° = p{,}[/latex] then [latex]\sin \underline{\qquad} = p{.}[/latex]
• If [latex]\sin 59° =w{,}[/latex] then [latex]\cos \underline{\qquad} = w{.}[/latex]
• If [latex]\sin \phi = \dfrac{5}{13}[/latex] and [latex]\cos \phi = \dfrac{12}{13}{,}[/latex] then [latex]\tan \phi =\underline{\qquad} {.}[/latex]
• If [latex]\cos \beta = \dfrac{1}{\sqrt{10}}{,}[/latex] and [latex]\sin \beta = \dfrac{3}{\sqrt{10}}{,}[/latex] then [latex]\tan \beta = \underline{\qquad}{.}[/latex]
• If [latex]\tan B = \dfrac{2}{\sqrt{5}}[/latex] and [latex]\cos B = \dfrac{\sqrt{5}}{3}{,}[/latex] then [latex]\sin B =\underline{\qquad}{.}[/latex]
• If [latex]\sin W = \sqrt{\dfrac{3}{7}}[/latex] and [latex]\tan W = \dfrac{\sqrt{3}}{2}{,}[/latex] then [latex]\cos W =\underline{\qquad}{.}[/latex]
• If [latex]\cos \theta = \dfrac{2}{\sqrt{10}}[/latex] and [latex]\sin \theta = \sqrt{\dfrac{3}{5}}{,}[/latex] then [latex]\tan \theta = \underline{\qquad}{.}[/latex]
• If [latex]\sin \alpha = \dfrac{\sqrt{2}}{4}{,}[/latex] and [latex]\cos \alpha =\dfrac{\sqrt{14}}{4}{,}[/latex] then [latex]\tan \alpha = \underline{\qquad}{.}[/latex]
• If [latex]\tan A = \dfrac{\sqrt{7}}{3}[/latex] and [latex]\cos A = \dfrac{3}{4}{,}[/latex] then [latex]\sin A =\underline{\qquad}{.}[/latex]
• If [latex]\sin V = \sqrt{\dfrac{10}{5}}[/latex] and [latex]\tan V = \dfrac{2}{5}{,}[/latex] then [latex]\cos V =\underline{\qquad}{.}[/latex]

Explain why the cosine of a [latex]73°[/latex] angle is always the same, no matter what size triangle the angle is in. Illustrate your explanation with a sketch.

• If you plotted the points in your table, would they lie on a straight line? Why or why not?
• What is the slope of the line through the origin and point [latex]P{?}[/latex]
• What is the tangent of the angle [latex]\theta{?}[/latex]
• On the same grid, sketch an angle whose tangent is [latex]\dfrac{8}{5}.[/latex]
• Use your calculator to complete the table. Round your answers to hundredths.
• Use the values of tan [latex]\theta[/latex] to sketch all the angles listed in the table. Locate the vertex of each angle at the origin and the initial side along the positive [latex]x[/latex]-axis.

Trigonometry Copyright © 2024 by Bimal Kunwor; Donna Densmore; Jared Eusea; and Yi Zhen. All Rights Reserved.

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Unit 9: Right triangles & trigonometry

About this unit, pythagorean theorem.

• Getting ready for right triangles and trigonometry (Opens a modal)
• Pythagorean theorem in 3D (Opens a modal)
• Pythagorean theorem with isosceles triangle (Opens a modal)
• Multi-step word problem with Pythagorean theorem (Opens a modal)
• Pythagorean theorem in 3D Get 3 of 4 questions to level up!
• Pythagorean theorem challenge Get 3 of 4 questions to level up!

Pythagorean theorem proofs

• Garfield's proof of the Pythagorean theorem (Opens a modal)
• Bhaskara's proof of the Pythagorean theorem (Opens a modal)
• Pythagorean theorem proof using similarity (Opens a modal)
• Another Pythagorean theorem proof (Opens a modal)

Special right triangles

• Special right triangles proof (part 1) (Opens a modal)
• Special right triangles proof (part 2) (Opens a modal)
• 30-60-90 triangle example problem (Opens a modal)
• Area of a regular hexagon (Opens a modal)
• Special right triangles review (Opens a modal)
• Special right triangles Get 3 of 4 questions to level up!

Ratios in right triangles

• Hypotenuse, opposite, and adjacent (Opens a modal)
• Side ratios in right triangles as a function of the angles (Opens a modal)
• Using similarity to estimate ratio between side lengths (Opens a modal)
• Using right triangle ratios to approximate angle measure (Opens a modal)
• Use ratios in right triangles Get 3 of 4 questions to level up!

Introduction to the trigonometric ratios

• Triangle similarity & the trigonometric ratios (Opens a modal)
• Trigonometric ratios in right triangles (Opens a modal)
• Trigonometric ratios in right triangles Get 3 of 4 questions to level up!

Solving for a side in a right triangle using the trigonometric ratios

• Solving for a side in right triangles with trigonometry (Opens a modal)
• Solve for a side in right triangles Get 3 of 4 questions to level up!

Solving for an angle in a right triangle using the trigonometric ratios

• Intro to inverse trig functions (Opens a modal)
• Solve for an angle in right triangles Get 3 of 4 questions to level up!

Sine & cosine of complementary angles

• Sine & cosine of complementary angles (Opens a modal)
• Using complementary angles (Opens a modal)
• Trig word problem: complementary angles (Opens a modal)
• Trig challenge problem: trig values & side ratios (Opens a modal)
• Relate ratios in right triangles Get 3 of 4 questions to level up!

Modeling with right triangles

• Right triangle word problem (Opens a modal)
• Angles of elevation and depression (Opens a modal)
• Right triangle trigonometry review (Opens a modal)
• Right triangle trigonometry word problems Get 3 of 4 questions to level up!

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11.1: The Trigonometric Ratios

• Last updated
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• Page ID 122907

• Katherine Yoshiwara
• Los Angeles Pierce College

2.1 Side and Angle Relationships

Homework 2.1.

1. The sum of the angles is not \(180^{\circ}\)

3. The exterior angle is not equal to the sum of the opposite interior angles.

5. The sum of the acute angles is not \(90^{\circ}\)

7. The largest side is not opposite the largest angle.

9. The Pythagorean theorem is not satisfied.

11. \(5^2+12^2=13^2\), but the angle opposite the side of length 13 is \(85^{\circ}\).

13. \(4 < x < 16\)

15. \(0 < x < 16\)

19. \(6\sqrt{2}\)in

21. \(w = 6\sqrt{10}\)in

25. \(\sqrt{3}\)

33. The distance from \((0,0)\) to \((3,3)\) is \(3 \sqrt{2}\), and the distance from \((3,3)\) to \((6,0)\) is also \(3 \sqrt{2}\), so the triangle is isosceles. The distance from \((0,0)\) to \((6,0)\) is 6 , and \((3 \sqrt{2})^2+(3 \sqrt{2})^2=6^2\) so the triangle is a right triangle.

37. \(\alpha=30^{\circ}, \beta=60^{\circ}, h=\sqrt{3}\)

39. \(8 \sqrt{3}\) in

a \((-1,0)\) and \((1,0) ; 2\)

b \(\sqrt{(p+1)^2+q^2}\) and \(\sqrt{(p-1)^2+q^2}\)

\begin{aligned} \left(\sqrt{(p+1)^2+q^2}\right)^2 & +\left(\sqrt{(p-1)^2+q^2}\right)^2 \\ & =p^2+2 p+1+q^2+p^2-2 p+1+q^2 \\ & =2 p^2+2+2 q^2=2+2\left(p^2+q^2\right) \\ & =2+2(1)=4 \end{aligned}

2.2 Right Triangle Trigonometry

Homework 2.2.

a \(4\sqrt{13} \approx 14.42\)

b \(\sin \theta = 0.5547, \cos \theta = 0.8321, \tan \theta = 0.6667

a \(4\sqrt{15} \approx 15.49\)

b \(\sin \theta = 0.9682, \cos \theta = 0.2500, \tan \theta = 3.8730\)

a \(2\sqrt{67} \approx 15.49\)

b \(\sin \theta = 0.2116, \cos \theta = 0.9774, \tan \theta = 0.2165\)

(Answers may vary)

b \(\tan 54.8^{\circ} = \dfrac{h}{20}, 170.1\) yd

b \(\tan 36.2^{\circ} = \dfrac{260}{d}, 355.2\) ft

b \(\sin 48^{\circ} = \dfrac{a}{1500}, 1114.7\)m

b \(\cos 38^{\circ} = \dfrac{1800}{x}, 2284.2\)m

35. \(x = \dfrac{82}{\tan \theta}\)

37. \(x = 11 \sin \theta\)

39. \(x = \dfrac{9}{\cos \theta}\)

41. \(36 \sin 25^{\circ} \approx 15.21\)

43. \(46 \sin 20^{\circ} \approx 15.73\)

45. \(12 \sin 40^{\circ} \approx 7.71\)

a \(\theta\) and \(\phi\) are complements.

b \(\sin \theta=\cos \phi\) and \(\cos \theta=\sin \phi\). The side opposite \(\theta\) is the side adjacent to \(\phi\), and vice versa.

a As \(\theta\) increases, \(\tan \theta\) increases also. The side opposite \(\theta\) increases in length while the side adjacent to \(\theta\) remains fixed.

b As \(\theta\) increases, \(\cos \theta\) decreases. The side adjacent to \(\theta\) remains fixed while the hypotenuse increases in length.

55. As \(\theta\) decreases toward \(0^{\circ}\), the side opposite \(\theta\) approaches a length of 0, so \(\sin \theta\) approaches 0. But as \(\theta\) increases toward \(90^{\circ}\), the length of the side opposite \(\theta\) approaches the length of the hypotenuse, so \(\sin \theta\) approaches 1.

57. The triangle is not a right triangle.

59. \(\dfrac{21}{20}\) is the ratio of hypotenuse to the adjacent side, which is the reciprocal of \(\cos \theta\).

a \(0.2358\)

c \(48^{\circ}\)

d \(77^{\circ}\)

a \(\dfrac{5}{12}\)

c \(\dfrac{2}{3}\)

d \(\dfrac{2}{\sqrt{7}}\)

65. Although the triangles may differ in size, the ratio of the side adjacent to the angle to the hypotenuse of the triangle remains the same because the triangles would all be similar, and hence corresponding sides are proportional.

a \(\dfrac{2}{3}\)

b \(\dfrac{2}{3}\)

2.3 Solving Right Triangles

Homework 2.3.

1. \(A = 61^{\circ}, a = 25.26, c = 28.88\)

3. \(A = 68^{\circ}, a = 0.93, b = 0.37\)

b \(B = 48^{\circ}, a = 17.4, b = 19.3\)

b \(A=57^{\circ}, b=194.4, c=357.7\)

b \(B=78^{\circ}, b=18.8, c=19.2\)

• Solve \(\sin 53.7^{\circ} = \dfrac{8.2}{c}\) for \(c\).
• Solve \(\tan 53.7^{\circ} = \dfrac{8.2}{a}\) for \(a\).
• Subtract \(53.7^{\circ}\) from \(90^{\circ}\) to find \(A\).

• Solve \(\cos 25^{\circ} = \dfrac{40}{c}\) for \(c\).
• Solve \(\tan 25^{\circ} = \dfrac{a}{40}\) for \(a\).
• Subtract \(25^{\circ}\) from \(90^{\circ}\) to find \(B\).

• Solve \(\sin 64.5^{\circ} = \dfrac{a}{24}\) for \(a\).
• Solve \(\cos 64.5^{\circ} = \dfrac{b}{24}\) for \(b\).
• Subtract \(64.5^{\circ}\) from \(90^{\circ}\) to find \(B\).

17. \(74.2^{\circ}\)

19. \(56.4^{\circ}\)

21. \(66.0^{\circ}\)

23. \(11.5^{\circ}\)

25. \(56.3^{\circ}\)

27. \(73.5^{\circ}\)

29. \(\cos 15^{\circ} = 0.9659\) and \(\cos ^{-1} 0.9659 = 15^{\circ}\)

31. \(\tan 65^{\circ} = 2.1445\) and \(\tan ^{-1} 2.1445 = 65^{\circ}\)

33. \(\sin ^{-1} (0.6) \approx 36.87^{\circ}\) is the angle whose sine is 0.6. \((\sin 6^{\circ})^{-1} \approx 9.5668\) is the reciprocal of \(\sin 6^{\circ}\).

b \(\sin \theta = \dfrac{1806}{3(2458)}, 14.6^{\circ}\)

b \(\tan \theta=\dfrac{32}{10}, 72.6^{\circ}\)

b \(c=10 \sqrt{10} \approx 31.6, A \approx 34.7^{\circ}, B \approx 55.3^{\circ}\)

b \(a=\sqrt{256.28} \approx 16.0, A \approx 56.5^{\circ}, B \approx 33.5^{\circ}\)

b \(\tan ^{-1}\left(\dfrac{26}{30}\right) \approx 40.9^{\circ}, \quad 91 \sqrt{1676} \approx 3612.6 \mathrm{~cm}\)

49. (a) and (b)

51. (a) and (d)

53. \(\dfrac{\sqrt{3}}{2} \approx 0.8660\)

55. \(\dfrac{1}{\sqrt{3}} = \dfrac{\sqrt{3}}{3} \approx 0.5774\)

63. \(a=3 \sqrt{3}, b=3, B=30^{\circ}\)

65. \(a=b=4 \sqrt{2}, B=45^{\circ}\)

67. \(e=4, f=4 \sqrt{3}, F=120^{\circ}\)

69. \(d=2 \sqrt{3}, e=2 \sqrt{2}, f=\sqrt{2}+\sqrt{6}, F=75^{\circ}\)

71. \(\a=20, b=20, c=20 \sqrt{2})

a \(32 \sqrt{3} \mathrm{~cm}\)

b \(128 \sqrt{3} \mathrm{sq} \mathrm{cm}\)

a \(10 \mathrm{sq} \mathrm{cm}\)

b \(10 \sqrt{2} \mathrm{sq} \mathrm{cm}\)

c \(10 \sqrt{3} \mathrm{sq} \mathrm{cm}\)

2.4 Chapter 2 Summary and Review

Chapter 2 review problems.

1. If \(C>93^{\circ}\), then \(A+B+C>180^{\circ}\)

3. If \(A<B<58^{\circ}\), then \(A+B+C<180^{\circ}\)

5. If \(C>50^{\circ}\), then \(A+B+C>180^{\circ}\)

9. \(a = 97\)

11. \(c = 52\)

15. \(\theta=35.26^{\circ}\)

17. No. \(a=6, c=10\) or \(a=9, c=15\)

a \(w=86.05\)

b \(\sin \theta=0.7786, \quad \cos \theta=0.6275, \quad \tan \theta=1.2407\)

a \(y=16.52\)

b \(\sin \theta=0.6957, \quad \cos \theta=0.7184, \quad \tan \theta=0.9684\)

23. \(a = 7.89\)

25. \(x = 3.57\)

27. \(b = 156.95\)

29. \(A=30^{\circ}, a=\dfrac{23 \sqrt{3}}{3}, c=\dfrac{46 \sqrt{3}}{3}\)

31. \(F=105^{\circ}, d=10 \sqrt{2}, e=20, f=10+10 \sqrt{3}\)

35. 43.30 cm

37. 15.92 m

39. \(114.02 \mathrm{ft}, 37.87^{\circ}\)

a \(60.26^{\circ}\)

b \(60.26^{\circ}\)

c \(m=\dfrac{7}{4}=\tan \theta\)

b \(b-a,(b-a)^2\)

c \(\dfrac{1}{2} a b\)

d \(4\left(\dfrac{1}{2} a b\right)+(a-b)^2=2 a b+b^2-2 a b+a^2=a^2+b^2\)